Two simple harmonic motions are represented by $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$ and $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$. The ratio of their amplitudes is

The $S.H.M.$ of a particle is given by the equation $y = 3\sin \omega t + 4\cos \omega t$. The amplitude is

$A$ particle is subjected to two mutually perpendicular simple harmonic motions such that its $x$ and $y$ coordinates are given by:
$x = 2 \sin \omega t$
$y = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$
The path of the particle will be:

If the displacement $y$ (in $cm$) of a particle executing simple harmonic motion is given by the equation $y = 5 \sin(3 \pi t) + 5 \sqrt{3} \cos(3 \pi t)$,then the amplitude of the particle is

When two displacements represented by $y_1 = a \sin(\omega t)$ and $y_2 = b \cos(\omega t)$ are superimposed,the motion is

$A$ particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference $(\delta)$ between the two motions is